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Intrgration
∫ x e^x / (1+x)^2 dx what's the answer?? 更新: ∫xxxxx 更新 2: ∫ x e^x / (1+x)^2 dx 更新 3: wt's the answer? 更新 4: can u give more steps or explain at in integration by part applying this question, it is in row 2 to row i dont understand,...thks
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圖片參考：http://i388.photobucket.com/albums/oo325/loyitak1990/Nov09/Crazyint2.jpg aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa 2009-11-16 22:57:08 補充： This employs the integration by parts in row 1: ∫udv = uv - ∫vdu By putting v = 1/(x + 1) and u = xe^x, we have: uv - ∫vdu = xe^x/(x + 1) - ∫d(xe^x)/(x + 1) Then for d(xe^x), we can further divided it into: d(xe^x) = e^x dx + xd(e^x) = e^x dx + xe^xdx
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Intrgration
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發問:∫ x e^x / (1+x)^2 dx what's the answer?? 更新: ∫xxxxx 更新 2: ∫ x e^x / (1+x)^2 dx 更新 3: wt's the answer? 更新 4: can u give more steps or explain at in integration by part applying this question, it is in row 2 to row i dont understand,...thks
最佳解答:
圖片參考：http://i388.photobucket.com/albums/oo325/loyitak1990/Nov09/Crazyint2.jpg aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa 2009-11-16 22:57:08 補充： This employs the integration by parts in row 1: ∫udv = uv - ∫vdu By putting v = 1/(x + 1) and u = xe^x, we have: uv - ∫vdu = xe^x/(x + 1) - ∫d(xe^x)/(x + 1) Then for d(xe^x), we can further divided it into: d(xe^x) = e^x dx + xd(e^x) = e^x dx + xe^xdx
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